We know that for general modules over a commutative ring with $1$, you can't always extract a basis from a generating set.

This makes me think that maybe there should be free modules of infinite rank which could be finitely generated. Do such things exist?

  • $\begingroup$ This is one of the most posted question. Please use search and read this. You will find that the answer to your question is no. $\endgroup$
    – user26857
    Dec 4, 2012 at 0:02

1 Answer 1


Suppose we had such a horrible thing, a surjection: $$R^{\oplus n} \twoheadrightarrow R^{\oplus I}$$where $I$ is something infinite (or just finite and $> n$). Pick any maximal ideal of $R$ and tensor up with $R/\mathfrak{m}$, it's right exact so we still have: $$R^{\oplus n} \otimes R/\mathfrak{m} \twoheadrightarrow R^{\oplus I} \otimes R/\mathfrak{m}$$But these are isomorphic to: $$R/\mathfrak{m}^{\oplus n} \twoheadrightarrow R/\mathfrak{m}^{\oplus I}$$a surjection of vector spaces.

So, can't happen!

  • 2
    $\begingroup$ Yeah another way to see it would be to write each generator in a finite generating set as a linear combination of basis vectors (finitely many in each case). We then deduce that every element in $M$ is a finite $R$-linear sum over this finite set of independent vectors; thus we've shown our basis was finite to begin with. $\endgroup$
    – JessicaB
    Dec 6, 2012 at 22:27

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